Answer
$f'\left( x \right) = \frac{{2x + 3}}{{{x^2} + 3x + 5}}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {{x^2} + 3x + 5} \right) \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {{x^2} + 3x + 5} \right)} \right] \cr
& {\text{Use the formula }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ }} {\text{ }} \cr
& {\text{Let }}u = {x^2} + 3x + 5,{\text{ then}} \cr
& f'\left( x \right) = \frac{1}{{{x^2} + 3x + 5}}\frac{d}{{dx}}\left[ {{x^2} + 3x + 5} \right] \cr
& f'\left( x \right) = \frac{1}{{{x^2} + 3x + 5}}\left( {2x + 3} \right) \cr
& {\text{simplify}} \cr
& f'\left( x \right) = \frac{{2x + 3}}{{{x^2} + 3x + 5}} \cr} $$
